1
$\begingroup$

Evaluate $\lim\limits_{x\rightarrow \infty}(1+\frac{1}{\sqrt{x}})^{\sqrt{x}}$.


Can I get some help? I am thinking that the limit does not exist. If you approach it from the left and then from the right, I think that the limits do not equal each other. I also suspect that we are dealing with Euler's limit, i.e.

The limits $\lim\limits_{x\rightarrow \infty +} (1+\frac{1}{x})^x$ and $\lim\limits_{x\rightarrow \infty -} (1+\frac{1}{x})^x$ exist and both equal $e$

Otherwise, I have little intuition to go from.

$\endgroup$
  • $\begingroup$ I edited that out, assuming it was a mistake. Apologies OP if that is incorrect. $\endgroup$ – James Apr 21 '15 at 14:26
  • $\begingroup$ $\sqrt{x}$ does not exist when $x<0$ and $+\infty$ can be approached only from the left. $\endgroup$ – Jack D'Aurizio Apr 21 '15 at 14:27
  • $\begingroup$ What does $\infty+$ and $\infty-$ mean? Do you mean $+\infty, -\infty$? $\endgroup$ – GFauxPas Apr 21 '15 at 14:29
  • 1
    $\begingroup$ Note that in real analysis $\infty$ is not a real point. We merely use the notation $lim_{x \to \infty} f(x) = L$ to mean $\forall \epsilon > 0 \exists M \in \mathbb{R}: \forall x \in \mathbb{R}\ x > M \implies | f(x) - L | < \epsilon$. This differs from the usual limit in that, instead of getting closer and closer to some value $L$ as we approach a point $a$, we now get closer and closer to some value $L$ as we make $x$ bigger and bigger. Consequently left- and right-limits do not exist for $\infty$. $\endgroup$ – Eric Spreen Apr 21 '15 at 14:38
  • $\begingroup$ In many books (imo, in most of them), writing the limit when $\;x\to\infty\;$ means $\;x\;$ approaches plus infinity , and there's only one way to approach this: from the left. For minus infinity we write $\;x\to -\infty\;$ $\endgroup$ – Timbuc Apr 21 '15 at 14:39
1
$\begingroup$

When $\;x\to\infty\;$ we can assume $\;x>0\;$ when doing the limit, so now simply make a substitution:

$$x\leftrightarrow y^2\;\implies\;\;x\to\infty\iff y\to\infty$$

and your limit becomes

$$\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$$

For negative $\;x$'s $\;\sqrt x\;$ isn't defined and thus also the limit of your expression isn't when $\;x\to -\infty\;$

$\endgroup$
0
$\begingroup$

Note that we can write

\begin{equation*} \lim_{x\to\infty} e^{(\sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} = e^{(\lim_{x\to\infty} \sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} =e^{\left(\lim_{x\to\infty}\frac{\ln(1+\frac{1}{\sqrt{x}})}{\frac{1}{\sqrt{x}}}\right)}. \end{equation*}

Applying L'Hopital's rule rule gives

\begin{equation*} e^{\left(\lim_{x\to\infty}\frac{1}{1+\frac{1}{\sqrt{x}}}\right)}=e. \end{equation*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.